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I'm having trouble doing the Sum-or-product of this truth table and then simplifying the logic expression.

For the SOP i got Y= A’.B’.C’.D’+ A.B’.C’.D’+ A’.B.C’.D’+ A’.B’.C.D’+ A’.B’.C’.D+ A.B.C’.D’+ A’.B.C.D’+ A’.B.C.D’+ A’.B.C’.D+ A’.B’.C.D+ A.B.C’.D+ A’.B.C.D+ A’.B’.C.D’+ A.B.C.D’+ A.B’.C.D

but I don't know where to go from here

enter image description here

Anarchy
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  • have idea about Karnaugh map/logic simplification. https://en.wikipedia.org/wiki/Karnaugh_map – user19579 Nov 04 '20 at 04:34
  • Why do you need more than 2 input bits to define 4 output states? – Tony Stewart EE75 Nov 04 '20 at 04:52
  • It seems , if I presume what you are doing, you only need 2 inputs. Do you want this to be a quadrature detector with wind direction but with a low pass filter to rotate the outputs in the shortest direction but averaged over time? That requires a state sequence not just combinational logic. https://i.stack.imgur.com/XUqrG.png Redefine your problem better – Tony Stewart EE75 Nov 04 '20 at 05:08
  • It's actually 4 inputs with 2 outputs, sorry if I didn't make it clear – Anarchy Nov 04 '20 at 05:09
  • That part is clear, but I guess 2 inputs are redundant – Tony Stewart EE75 Nov 04 '20 at 05:10
  • look at my table again E-W=A' and N-S=B' – Tony Stewart EE75 Nov 04 '20 at 05:12
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    @Anarchy You only show one Y equation. If you have two outputs then there should be two SOP equations. Yes? – jonk Nov 04 '20 at 05:26
  • That's what i thought, but the question after the one asking for the sop of the truth table is 'Simplify your logic expression obtained in the previous question", which implies that there are just one?? – Anarchy Nov 04 '20 at 05:58
  • @Anarchy No. You have two output columns, if I'm reading things right. That means two equations. To me, anyway. – jonk Nov 04 '20 at 06:15
  • I tried to find help online about what to do when there are two outputs in one equation but i did not find anything. Just one exemple that said to combine the results in the sop and just exclude the combinaisons where both outputs are low, which is what i tried to do :/ – Anarchy Nov 04 '20 at 06:19
  • A circuit with 2 outputs has two boolean expressions? – Anarchy Nov 04 '20 at 06:25
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    @Anarchy Yes. Of course. Each equation accounts for one of the outputs. Is that difficult to follow why? Treat each output column as a separate table. Doesn't that make sense? – jonk Nov 04 '20 at 06:28

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If you haven't yet gathered the idea of k-maps, you should work on that. In your example, here are the k-maps I see:

$$\begin{array}{rl} \begin{smallmatrix}\begin{array}{r|cccc} E-W&\overline{B}\:\overline{A}&\overline{B}\: A&B \:A&B \:\overline{A}\\ \hline \overline{D}\:\overline{C}&1&1&1&1\\ \overline{D}\:C &1&0&0&1\\ D\: C &1&0&0&1\\ D\:\overline{C} &1&1&1&1 \end{array}\end{smallmatrix} & \begin{smallmatrix}\begin{array}{r|cccc} N-S&\overline{B}\:\overline{A}&\overline{B}\: A&B \:A&B \:\overline{A}\\ \hline \overline{D}\:\overline{C}&0&1&1&0\\ \overline{D}\:C&1&1&1&1\\ D\: C&1&1&0&0\\ D\:\overline{C}&1&1&0&0 \end{array}\end{smallmatrix} \end{array}$$

Read the above, carefully. Does this make any sense to you?

jonk
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